Bài tập 25 trang 162 SGK Toán 12 NC

a) Đặt \(\left\{ \begin{array}{l}
u = x\\
dv = \cos 2xdx
\end{array} \right. \)

\(\Rightarrow \left\{ \begin{array}{l}
du = dx\\
v = \frac{1}{2}\sin 2x
\end{array} \right.\)

Do đó: 

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\int\limits_0^{\frac{\pi }{4}} x \cos 2xdx\\
 = \left. {\frac{1}{2}x\sin 2x} \right|_0^{\frac{\pi }{4}} - \frac{1}{2}\int\limits_0^{\frac{\pi }{4}} {\sin 2x} dx
\end{array}\\
\begin{array}{l}
 = \frac{\pi }{8} + \left. {\frac{1}{4}\cos 2x} \right|_0^{\frac{\pi }{4}}\\
 = \frac{\pi }{8} + \frac{1}{4}\left( { - 1} \right) = \frac{\pi }{8} - \frac{1}{4}
\end{array}
\end{array}\)

b) Đặt \(u = \ln (2 - x) \Rightarrow du = \frac{{ - 1}}{{2 - x}}dx\)

\(\begin{array}{l}
\int\limits_0^1 {\frac{{\ln \left( {2 - x} \right)}}{{2 - x}}} dx =  - \int\limits_{\ln 2}^0 u du\\
 = \int\limits_0^{\ln 2} u du = \left. {\frac{{{u^2}}}{2}} \right|_0^{\ln 2} = \frac{1}{2}{\left( {\ln 2} \right)^2}
\end{array}\)

c) Đặt \(\left\{ \begin{array}{l}
u = {x^2}\\
dv = \cos xdx
\end{array} \right. \)

\(\Rightarrow \left\{ \begin{array}{l}
du = 2xdx\\
v = \sin x
\end{array} \right.\)

Do đó: 

\(\begin{array}{l}
I = \int\limits_0^{\frac{\pi }{2}} {{x^2}} \cos 2xdx\\
 = \left. {{x^2}{\rm{sinx}}} \right|_0^{\frac{\pi }{2}} - 2\int\limits_0^{\frac{\pi }{2}} x sinxdx\\
 = \frac{{{\pi ^2}}}{4} - 2{I_1}
\end{array}\)

Với \({I_1} = \int \limits_0^{\frac{\pi }{2}} x\sin xdx\)

Đặt \(\left\{ {\begin{array}{*{20}{l}}
{u = x}\\
{dv = \sin xdx}
\end{array}} \right. \Rightarrow \left\{ \begin{array}{l}
du = dx\\
v =  - \cos x
\end{array} \right.\)

Do đó:

\(\begin{array}{l}
{I_1} = \left. { - x\cos x} \right|_0^{\frac{\pi }{2}} + \int\limits_0^{\frac{\pi }{2}} {\cos } xdx\\
 = \left. {\sin x} \right|_0^{\frac{\pi }{2}} = 1
\end{array}\)

Vậy \(I = \frac{{{\pi ^2}}}{4} - 2\)

d) Đặt

\(\begin{array}{l}
u = \sqrt {{x^3} + 1}  \Rightarrow {u^2} = {x^3} + 1\\
 \Rightarrow 2udu = 3{x^2}dx\\
 \Rightarrow {x^2}dx = \frac{2}{3}udu
\end{array}\)

\(\begin{array}{l}
\int\limits_0^1 {{x^2}} \sqrt {{x^3} + 1} dx = \frac{2}{3}\int\limits_1^{\sqrt 2 } {{u^2}} du\\
 = \left. {\frac{{2{u^3}}}{9}} \right|_1^{\sqrt 2 } = \frac{2}{9}\left( {2\sqrt 2 0 - 1} \right)
\end{array}\)

e) Đặt \(\left\{ \begin{array}{l}
u = \ln x\\
dv = {x^2}dx
\end{array} \right. \)

\(\Rightarrow \left\{ \begin{array}{l}
du = \frac{{dx}}{2}\\
v = \frac{{{x^3}}}{3}
\end{array} \right.\)

Do đó:

\(\begin{array}{l}
\int\limits_1^e {{x^2}} \ln xdx\\
 = \left. {\frac{{{x^3}}}{3}\ln x} \right|_1^e - \frac{1}{3}\int\limits_0^e {{x^2}} dx\\
 = \left. {\frac{{{e^3}}}{3} - \frac{1}{9}{x^3}} \right|_1^e = \frac{{2{e^3} + 1}}{9}
\end{array}\)

-- Mod Toán 12 HỌC247